write 7 test. Please.

Can someone send test 6 please? Thanks

Edited by author 26.11.2019 11:33

I used sqrt-decomposition and got AC in 358ms

My idea was like this:
1. Keep an array of the elements we have so far
2. For each sqrt-size block of the array compute the GCD
3. To insert an element: append it to the array and update its block
4. To remove an element: swap the element to be deleted with last element (I used a set of pairs<value, index> to get them quickly) and update their blocks

Insert (sqrt n)
Delete (sqrt n)
O(q sqrt n) where n is number of elements in the array

please can anyone tell me what is the test of 5.

#include <iostream>
#include <string>
using namespace std;
int main()
{
    int n,g=2;
    cin >> n;
    for(int i = 0; i < n; i++)
    {
        string a;
        cin >> a;
        if(a.length() > 4 && a.substr(a.length()-4,a.length()-1) == "+one")
            g+=2;
        else
            g++;
    }
    if(g == 13)
        cout << 1400;
    else
        cout << g*100;
}

Can you send me test 11?

I can't understand what's wrong with my code
is my algorithm wrong? or what?
my code:

#include <stdio.h>
#include <math.h>
double x[102],y[102];

double dist(int i, int j){
    int d1=abs(x[i]-x[j]),d2=abs(y[i]-y[j]);
    if(d1==0)
        return (d2);
    else if(d2==0)
      return (d1);
    else
      return sqrt(d1*d1 + d2*d2);
}

int main(){
    int i;
    double n,s=0,r,pi=2*acos(0);
  scanf("%Lf %Lf",&n,&r);
  for(i=0;i<n;i++)
      scanf("%Lf %Lf",&x[i],&y[i]);
    for(i=1;i<n;i++)
        s+=dist(i,i-1);
    s+=dist(0,n-1);
    s+=2*pi*r;
    printf("%.2Lf\n",s);
  return 0;
}

if someone can help me I would appreciate very much
thanks!

Edited by author 17.11.2019 11:13

Can anyone help me? I have no idea about how this test looks like.

Any hints or tests?

So, when user starts looking for words, my program immediately writes the answer. Maybe that's why it isn't correct. For example:
user -> 2
user -> ab
program -> abc
user -> xz
program -> xzy

We can qsort(n) and enumeration to solve it.
Notice that n is a prime.There are 2500 primes under 32767.
Enumeration can solve this problem.
And every ask have mostly 2 answers.最多2个解

I wrote the solution and had the correct answers on test cases, but i have wa1. Can someone help me or give advice?

This is my solution:

#include <iostream>
#include <iomanip>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <set>
#include <map>
#include <vector>
#include <string>
#include <cstring>
#include <cstdlib>

using namespace std;
#define task


void print(int array[], int n) {
     for (int i = 0; i < n; i++) {
          cout << array[i] << ' ';
     }
}

void optimization() {
     cin.tie(nullptr);
     ios_base::sync_with_stdio(false);
}

int factorial(int x) {
     int a = 1;
     for (int i = 2; i < x + 1; i++) {
          a *= i;
     }
     return a;
}


#ifdef task

int main() {
     optimization();

     int n, m;
     cin >> n >> m;
     int **field = new int *[n];
     for (int i = 0; i < n; i++) {
          field[i] = new int[m];
     }

     for (int i = 0; i < n; i++) {
          for (int j = 0; j < m; j++) {
               cin >> field[i][j];
          }
     }
     int q;
     int k;
     int x = 0; int y = 0;
     int max_drop = -1;
     int part_sum = 0;
     int move_to_right = -1;
     cin >> q;
     for (int quest = 0; quest < q; quest++) {
          cin >> k;
          for (int i = 0; i < n; i++) {
               for (int j = 0; j < m; j++) {
                    move_to_right = k + 1;
                    part_sum = 0;
                    for (int down = i; down < i + k + 1; down++) {
                         for (int right = j; right < j + move_to_right; right++) {
                              if (down < n && right < m) {
                                   part_sum += field[down][right];
                              }
                         }
                         move_to_right--;
                    }

                    if (part_sum > max_drop) {
                         max_drop = part_sum;
                         x = i + 1;
                         y = j + 1;
                    }
               }
          }

          cout << max_drop << ' ' << x << ' ' << y << endl;
          max_drop = -1;

     }

     return 0;
}

#endif

Edited by author 14.11.2019 22:47

In the example 2 2 3, z==5. Why do everybody say that z==2?

есть ли способы на питоне обойти прожорливость памяти?

some Test Please !!!